Mg(s) + 2 HCl(aq) --> MgCl2(aq) + H2 (g) DeltaH= -4.6x10^2 kJ

Question

How much heat is evolved when 0.598 g of Mg are dissolved in excess HCl?

I'm not sure how to go about this problem. What I thought to do was:

0.598 g x 1 mol Mg/24.31 g Mg x 2 mol HCl/ 1 mol Mg x 36.46 g HCl/ 1 mol HCl = 1.79 g HCl
and from here I am not sure what to do. Somehow I need to incorporate the DeltaH into a formula, I believe.

Answer 1

Your approach is to perform a stoichiometry calculation to find the grams HCl but that isn't what the problem is asking. Try something like this.
You know dissolving 1 mole Mg (24.3 g) in HCl produces 4.6E2 kJ. So how much heat will just 0.598 g Mg evolve?
That will be
4.6E2 kJ x 0.598/24.3 = ??

Answer 2

So the grams must cancel each other out when we divide and it leaves us with just kJ? So -11.3 kJ

Answer 3

Yes, but I notice there are only two significant figures in 4.6E2; therefore, the answer should not contain more than two. I would round 11.3 kJ to 11 kJ energy released (-11 kJ for delta H).

Answer 4

Oh that's right, thank you!